A magnitude of vector `vecA,vecB` and `vecC` are respectively `12, 5` and `13` units and `vecA+vecB=vecC` then the angle between `vecA` and `vecB` is

Options

A. 0
B. `pi//2`
C. `pi//4`
D. `pi`

Explanation

To find the angle between the vectors \(\vec{A}\) and \(\vec{B}\), we can use the information given in the problem along with the properties of vectors.

### Step-by-Step Solution:

1. **Identify the Magnitudes of the Vectors**:
- Given magnitudes are:
- \(|\vec{A}| = 12\)
- \(|\vec{B}| = 5\)
- \(|\vec{C}| = 13\)

2. **Use the Vector Addition Formula**:
- According to the problem, \(\vec{A} + \vec{B} = \vec{C}\).
- The magnitude of the resultant vector \(\vec{C}\) can be expressed using the formula:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos(\theta) \]
where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

3. **Substitute the Known Values**:
- Substitute the magnitudes into the equation:
\[ 13^2 = 12^2 + 5^2 + 2 \cdot 12 \cdot 5 \cdot \cos(\theta) \]
- This simplifies to:
\[ 169 = 144 + 25 + 120 \cos(\theta) \]

4. **Simplify the Equation**:
- Combine the known values:
\[ 169 = 169 + 120 \cos(\theta) \]
- Rearranging gives:
\[ 0 = 120 \cos(\theta) \]

5. **Solve for \(\cos(\theta)\)**:
- From the equation \(0 = 120 \cos(\theta)\), we find:
\[ \cos(\theta) = 0 \]

6. **Determine the Angle**:
- The cosine of an angle is zero at \(90^\circ\) (or \(\frac{\pi}{2}\) radians). Therefore:
\[ \theta = 90^\circ \]

### Conclusion:
The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\) or \(\frac{\pi}{2}\) radians.